Calculate the energy in the form of heat, in kJ, required to convert 225 grams of liquid water at 21.0 °C to steam at 115 °C. Heat of fusion = 0.333 kJ/g; heat of vaporization = 2.26 kJ/g; specific heat capacities: liquid water = 4.18 J/g·°C., steam = 1.92 J/g·°C.
A. 88.5 kJ
B. 158 kJ
C. 589 kJ
D. 596 kJ
E. 663 kJ
C. 589
To calculate the energy required to convert liquid water at 21.0 °C to steam at 115 °C, we need to consider three processes: heating the water from 21.0 °C to its boiling point, vaporizing the water at its boiling point, and heating the steam from its boiling point to 115 °C.
First, let's calculate the energy required to heat the water from 21.0 °C to its boiling point. We can use the formula:
Q = m * c * ΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
m = 225 g (mass of water)
c = 4.18 J/g·°C (specific heat capacity of liquid water)
ΔT = boiling point - initial temperature = 100 °C - 21.0 °C = 79.0 °C
Q1 = 225 g * 4.18 J/g·°C * 79.0 °C
Q1 = 74,493.5 J
Next, let's calculate the energy required to vaporize the water at its boiling point. We can use the formula:
Q = m * ΔHvap
where Q is the heat energy, m is the mass, and ΔHvap is the heat of vaporization.
Given:
m = 225 g (mass of water)
ΔHvap = 2.26 kJ/g
Q2 = 225 g * 2.26 kJ/g
Q2 = 508.5 kJ = 508,500 J
Finally, let's calculate the energy required to heat the steam from its boiling point to 115 °C. We can use the formula:
Q = m * c * ΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
m = 225 g (mass of water)
c = 1.92 J/g·°C (specific heat capacity of steam)
ΔT = final temperature - boiling point = 115 °C - 100 °C = 15 °C
Q3 = 225 g * 1.92 J/g·°C * 15 °C
Q3 = 6,480 J
Now, we can calculate the total energy required by adding the three quantities:
Total energy = Q1 + Q2 + Q3
Total energy = 74,493.5 J + 508,500 J + 6,480 J
Total energy = 589,473.5 J
To convert this energy to kilojoules (kJ), divide by 1000:
Total energy in kJ = 589,473.5 J / 1000
Total energy in kJ = 589.5 kJ
Therefore, the energy required to convert 225 grams of liquid water at 21.0 °C to steam at 115 °C is approximately 589.5 kJ.
The correct answer is C. 589 kJ.